1.07d Second derivatives: d^2y/dx^2 notation

10 \begin{figure}[h] \end{figure} Fig. 10 shows a solid cuboid with square base of side \(x \mathrm {~cm}\) and height \(h \mathrm {~cm}\). Its volume is \(120 \mathrm {~cm} ^ { 3 }\).

1. Find \(h\) in terms of \(x\). Hence show that the surface area, \(A \mathrm {~cm} ^ { 2 }\), of the cuboid is given by \(A = 2 x ^ { 2 } + \frac { 480 } { x }\).
2. Find \(\frac { \mathrm { d } A } { \mathrm {~d} x }\) and \(\frac { \mathrm { d } ^ { 2 } A } { \mathrm {~d} x ^ { 2 } }\).
3. Hence find the value of \(x\) which gives the minimum surface area. Find also the value of the surface area in this case.

4 A curve has equation \(y = x + \frac { 1 } { x }\).
Use calculus to show that the curve has a turning point at \(x = 1\).
Show also that this point is a minimum.

2 Given that \(y = 6 x ^ { \frac { 3 } { 2 } }\), find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) and \(\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }\).
Show, without using a calculator, that when \(x = 36\) the value of \(\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }\) is \(\frac { 3 } { 4 }\).

10 Fig. 10 shows the curve with equation \(y = x ^ { 2 } + \frac { 16 } { x }\). \begin{figure}[h] \end{figure}

1. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\).
2. Hence calculate the coordinates of the stationary point on the curve.
3. Find \(\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }\) and explain why this confirms that he stationary point is a minimum.
4. Using the trapezium rule with 4 intervals, estimate the area between the curve and the \(x\) axis between \(x = 2\) and \(x = 4\).
5. State, giving a reason, whether this estimate of the area under-estimates or over-estimates the true area beneath the curve.

10 Given tha \(y = 6 x ^ { \frac { 3 } { 2 } }\), find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) and \(\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }\).
Show, without using a calculator, that when \(x = 36\) the value of \(\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }\) is \(\frac { 3 } { 4 }\).

12 Given tha \(y = 6 x ^ { 3 } + \sqrt { x } + 3\), find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) and \(\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }\).

4 The equation of a curve is \(y = 9 x ^ { 2 } - x ^ { 4 }\).

1. Show that the curve meets the \(x\)-axis at the origin and at \(x = \pm a\), stating the value of \(a\).
2. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) and \(\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }\). Hence show that the origin is a minimum point on the curve. Find the \(x\)-coordinates of the maximum points.
3. Use calculus to find the area of the region bounded by the curve and the \(x\)-axis between \(x = 0\) and \(x = a\), using the value you found for \(a\) in part (i).

8 A curve has equation \(y = ( x + 2 ) \mathrm { e } ^ { - x }\).

1. Find the coordinates of the points where the curve cuts the axes.
2. Find the coordinates of the stationary point, S , on the curve.
3. By evaluating \(\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }\) at S , determine whether the stationary point is a maximum or a minimum.
4. Sketch the curve in the domain \(- 3 < x < 3\).
5. Find where the normal to the curve at the point \(( 0,2 )\) cuts the curve again.
6. Find the area of the region bounded by the curve, the \(x\)-axis and the lines \(x = 1\) and \(x = 3\).

1 Fig. 8 shows the line \(y = 1\) and the curve \(y = \mathrm { f } ( x )\), where \(\mathrm { f } ( x ) = \frac { ( x - 2 ) ^ { 2 } } { x }\). The curve touches the \(x\)-axis at \(\mathrm { P } ( 2,0 )\) and has another turning point at the point Q . \begin{figure}[h] \end{figure}

1. Show that \(\mathrm { f } ^ { \prime } ( x ) = 1 - \frac { 4 } { x ^ { 2 } }\), and find \(\mathrm { f } ^ { \prime \prime } ( x )\). Hence find the coordinates of Q and, using \(\mathrm { f } ^ { \prime \prime } ( x )\), verify that it is a maximum point.
2. Verify that the line \(y = 1\) meets the curve \(y = \mathrm { f } ( x )\) at the points with \(x\)-coordinates 1 and 4 . Hence find the exact area of the shaded region enclosed by the line and the curve. The curve \(y = \mathrm { f } ( x )\) is now transformed by a translation with vector \(\binom { - 1 } { - 1 }\). The resulting curve has equation \(y = \mathrm { g } ( x )\).
3. Show that \(\mathrm { g } ( x ) = \frac { x ^ { 2 } - 3 x } { x + 1 }\).
4. Without further calculation, write down the value of \(\int _ { 0 } ^ { 3 } \mathrm {~g} ( x ) \mathrm { d } x\), justifying your answer.

1 Fig. 8 shows the curve \(y = 3 \ln x + x - x ^ { 2 }\).
The curve crosses the \(x\)-axis at P and Q , and has a turning point at R . The \(x\)-coordinate of Q is approximately 2.05 . \begin{figure}[h] \end{figure}

1. Verify that the coordinates of P are \(( 1,0 )\).
2. Find the coordinates of R , giving the \(y\)-coordinate correct to 3 significant figures. Find \(\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }\), and use this to verify that R is a maximum point.
3. Find \(\int \ln x \mathrm {~d} x\). Hence calculate the area of the region enclosed by the curve and the \(x\)-axis between P and Q , giving your answer to 2 significant figures.

3 A curve is defined by the equation \(y = 2 x \ln ( 1 + x )\).

1. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) and hence verify that the origin is a stationary point of the curve.
2. Find \(\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }\), and use this to verify that the origin is a minimum point.
3. Using the substitution \(u = 1 + x\), show that \(\int \frac { x ^ { 2 } } { 1 + x } \mathrm {~d} x = \int \left( u - 2 + \frac { 1 } { u } \right) \mathrm { d } u\). Hence evaluate \(\int _ { 0 } ^ { 1 } \frac { x ^ { 2 } } { 1 + x } \mathrm {~d} x\), giving your answer in an exact form.
4. Using integration by parts and your answer to part (iii), evaluate \(\int _ { 0 } ^ { 1 } 2 x \ln ( 1 + x ) \mathrm { d } x\).

2

1. Given that \(y = \tan ^ { - 1 } x\), prove that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 1 } { 1 + x ^ { 2 } }\).
2. Verify that \(y = \tan ^ { - 1 } x\) satisfies the equation $$\left( 1 + x ^ { 2 } \right) \frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 2 x \frac { \mathrm {~d} y } { \mathrm {~d} x } = 0$$

9 Given that \(\mathrm { f } ( x ) = \frac { 1 } { x } - \sqrt { x } + 3\),

1. find \(\mathrm { f } ^ { \prime } ( x )\),
2. find \(\mathrm { f } ^ { \prime \prime } ( 4 )\).

6 Given that \(y = \frac { 5 } { x ^ { 2 } } - \frac { 1 } { 4 x } + x\), find

1. \(\frac { \mathrm { d } y } { \mathrm {~d} x }\),
2. \(\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }\).

6 Given that \(\mathrm { f } ( x ) = \frac { 4 } { x } - 3 x + 2\),

1. find \(\mathrm { f } ^ { \prime } ( x )\),
2. find \(\mathrm { f } ^ { \prime \prime } \left( \frac { 1 } { 2 } \right)\).

1 Given that \(y = x ^ { 5 } + \frac { 1 } { x ^ { 2 } }\), find

1. \(\frac { \mathrm { d } y } { \mathrm {~d} x }\),
2. \(\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }\).

5 The equation of a curve is \(y = x ^ { 2 } \ln ( 4 x - 3 )\). Find the exact value of \(\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }\) at the point on the curve for which \(x = 2\).

1 Given that \(y = 4 x ^ { 2 } \ln x\), find the value of \(\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }\) when \(x = \mathrm { e } ^ { 2 }\).

3 A curve \(C\) has equation \(\tan y = x\), for \(x > 0\).

1. Use implicit differentiation to show that $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } = - 2 x \left( \frac { \mathrm {~d} y } { \mathrm {~d} x } \right) ^ { 2 }$$
2. Hence find the value of \(\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }\) at the point \(\left( 1 , \frac { 1 } { 4 } \pi \right)\) on \(C\).

4 A curve \(C\) has equation \(x ^ { 3 } - 3 x y + y ^ { 2 } = 4\). Find the value of \(\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }\) at the point \(( 0,2 )\) of \(C\).

2 A curve has equation \(y = x ^ { 5 } - 5 x ^ { 4 }\).

1. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) and \(\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }\).
2. Verify that the curve has a stationary point when \(x = 4\).
3. Determine the nature of this stationary point.

1. A curve \(C\) has equation

$$y = x ^ { 2 } - 2 x - 24 \sqrt { x } , \quad x > 0$$

1. Find (i) \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) (ii) \(\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }\)
2. Verify that \(C\) has a stationary point when \(x = 4\)
3. Determine the nature of this stationary point, giving a reason for your answer.

1. (i) Use proof by exhaustion to show that for \(n \in \mathbb { N } , n \leqslant 4\)

$$( n + 1 ) ^ { 3 } > 3 ^ { n }$$ (ii) Given that \(m ^ { 3 } + 5\) is odd, use proof by contradiction to show, using algebra, that \(m\) is even.

1. $$y = 4 x ^ { 3 } - 7 x ^ { 2 } + 5 x - 10$$

1. Find in simplest form
   1. \(\frac { \mathrm { d } y } { \mathrm {~d} x }\)
   2. \(\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }\)
2. Hence find the exact value of \(x\) when \(\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } = 0\)

1. The curve \(C\) has equation

$$y = 5 x ^ { 4 } - 24 x ^ { 3 } + 42 x ^ { 2 } - 32 x + 11 \quad x \in \mathbb { R }$$

1. Find
   1. \(\frac { \mathrm { d } y } { \mathrm {~d} x }\)
   2. \(\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }\)
2. 1. Verify that \(C\) has a stationary point at \(x = 1\)
   2. Show that this stationary point is a point of inflection, giving reasons for your answer.